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a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
a: 7x+58=100
nên 7x=42
hay x=6
c: x-56:x=16
nên x-14=16
hay x=30
c)x - 56 : 4 = 16
x - 56 = 16 : 4
x- 56 = 4
x =4 + 56
x = 60
d)101 + (36 - 4x) = 105
(36- 4x ) = 105 - 101
36 - 4x = 4
4x = 36 - 4
4x = 32
x = 32:4
x = 8
d, \(=>\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4.\)
=> \(2x+7=4\)
=> 2x= -3
=> x=-3/2 . Vậy x=-3/2
e, => \(\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^2}{131}.\)
=> \(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}\left(1+5+5^2\right)}{131}\)
= > \(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
=> \(7^x=5^{2x}\)
Đến đoạn này là mik nghĩ không ra nhé
Cô làm tiếp giúp Linh Đan:
\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)
a) \(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^2-\left(x-4^4\right)=0\)
\(\Rightarrow\left(x-4\right)^2.\left[1-\left(x-4\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\\left(x-4\right)^2=1^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)
a) 2x + 23x = 625
x (2 + 23) = 625
x. 25 = 625
x = 625 : 25
x = 25
b) (2^2+4^2) x +2^4 x 5 * x = 100 (dấu* là dấu nhân vì mik không muốn trùng nhau nên viết vậy)
( 4 + 16 ) x + 2^4 x 5 * x = 100
20 * x + 16 x 5 * x =100
x ( 20 + 16 x 5 ) = 100
x ( 20 + 80 ) = 100
x * 100 =100
x=100 : 100
x= 1
c) ( 3x +1)^2 = 6^2+8^2
(3x +1)^2 = 36 +64
(3x +1)^2 =100
(3x +1)^2 = 10^2
3x + 1 =10
3x=10 - 1=9
x = 9 :3
x = 3
( d và e làm tương tự )
k mik nha!
a) |x-1|-2x=5
\(\Leftrightarrow\) x - 1 - 2x = 5 (đk: x\(\ge\)1 ) (1)
hoặc -x + 1 - 2x = 5 ( đk: x < 1) (2)
(1): \(\Leftrightarrow\) x - 2x = 5 + 1
\(\Leftrightarrow\) - x = 6
\(\Leftrightarrow\) x = -6 (ko thỏa mãn)
(2): \(\Leftrightarrow\) -x + 1 - 2x = 5
\(\Leftrightarrow\) -3x = 4
\(\Leftrightarrow\) x = \(\frac{-4}{3}\) (thỏa mãn)
Vậy: x = \(\frac{-4}{3}\)
b) |9-7x|=5x-3
\(\Leftrightarrow\) 9 - 7x = 5x - 3 (đk: \(\le\) \(\frac{9}{7}\) ) (1)
hoặc -9 + 7x = 5x - 3 (đk: x > \(\frac{9}{7}\) ) (2)
(1): \(\Leftrightarrow\) -7x-5x = -3-9 \(\Leftrightarrow\) -12x = -12 \(\Leftrightarrow\) x = 1 ( thỏa mãn)
(2): \(\Leftrightarrow\) 7x - 5x = -3 + 9 \(\Leftrightarrow\) 2x = 6 \(\Leftrightarrow\) x = 3 ( thỏa mãn)
Vậy: x= 1; 3
c) 5x+2 = 625
\(\Leftrightarrow\) 5x . 52 = 54
\(\Leftrightarrow\) 5x = \(\frac{5^4}{5^2}\)= 52
\(\Leftrightarrow\) x = 2
d) (2x - 3)2 = 36
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-3=6\left(đk:x\ge\frac{3}{2}\right)\text{}\Leftrightarrow x=4,5\left(TMĐK\right)\\2x-3=-6\left(đk:x< \frac{3}{2}\right)\Leftrightarrow x=-1,5\left(TMĐK\right)\end{matrix}\right.\)
Vậy: x = 4,5; -1,5
19 22 25 28 5(3x + 2) – 4(2x +3) x*(1 + 2x) 4(1 + x) – 3(2x-5) 4x–8(6) - X) 23/ ... 2x” – 4x + 3x – 6 = 2x” – X-6 (b) (x-3) = (x-3)(x-3) (c) (2x+y)(2x–y) = x* = x* – 3x ... (x - 6)” 7 (3x + 5)(x-6) 8 (8x + 2)(3x + 4) (4x – 1)(2x – 3) 10 (2x +5)* 11 (8x – 3)(2x + ... 27 (4x + 3y)(x + y) 28 (2x + 5)(5x – 2) (4x – 3y)(4x + y) 30 (7x + 2y)(3x + 4y) 24/ ...
\(a)=3x\cdot\left(2x-7-4x+5\right)=3x\cdot\left(-2x-2\right)=3x\cdot\left[-2\cdot\left(x+1\right)\right]\)
a: \(\Leftrightarrow2x+7=-4\)
=>2x=-11
hay x=-11/2
b: \(\Leftrightarrow\dfrac{7^x\cdot49+7^x\cdot7+7^x}{57}=\dfrac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Leftrightarrow7^x=5^{2x}\)
=>x=0